Mixing
Let $W_i$ be the null space of $(T-c_i I)$. If $T$ is diagonalizable, then prove that $sum_i dim(W_i) = dim...
$begingroup$
Let $T$ be a linear operator on a finite dimensional vector space $V$ and let $c_1,c_2ldots ,c_k$ be the distinct eigenvalues of T. Let $W_i$ be the null space of $(T-c_i I)$. If $T$ is diagonalizable, then prove that $$sum_i dim(W_i) = dim(V)$$
I am unable to connect things in order to logically think about this. How do I go about proving this?
linear-algebra vector-spaces linear-transformations
edited Jan 29 at 7:17
Chinnapparaj R
5,8602928
asked Jan 29 at 7:13
AbhayAbhay
3789
$endgroup$
add a comment |
$begingroup$
Let $T$ be a linear operator on a finite dimensional vector space $V$ and let $c_1,c_2ldots ,c_k$ be the distinct eigenvalues of T. Let $W_i$ be the null space of $(T-c_i I)$. If $T$ is diagonalizable, then prove that $$sum_i dim(W_i) = dim(V)$$
I am unable to connect things in order to logically think about this. How do I go about proving this?
linear-algebra vector-spaces linear-transformations
edited Jan 29 at 7:17
Chinnapparaj R
5,8602928
asked Jan 29 at 7:13
AbhayAbhay
3789
$endgroup$
$begingroup$
See proposition $5.21$ on page no:$101$ in this book!
$endgroup$
– Chinnapparaj R
Jan 29 at 7:36
add a comment |
$begingroup$
Let $T$ be a linear operator on a finite dimensional vector space $V$ and let $c_1,c_2ldots ,c_k$ be the distinct eigenvalues of T. Let $W_i$ be the null space of $(T-c_i I)$. If $T$ is diagonalizable, then prove that $$sum_i dim(W_i) = dim(V)$$
I am unable to connect things in order to logically think about this. How do I go about proving this?
linear-algebra vector-spaces linear-transformations
edited Jan 29 at 7:17
Chinnapparaj R
5,8602928
asked Jan 29 at 7:13
AbhayAbhay
3789
$endgroup$
Let $T$ be a linear operator on a finite dimensional vector space $V$ and let $c_1,c_2ldots ,c_k$ be the distinct eigenvalues of T. Let $W_i$ be the null space of $(T-c_i I)$. If $T$ is diagonalizable, then prove that $$sum_i dim(W_i) = dim(V)$$
I am unable to connect things in order to logically think about this. How do I go about proving this?
linear-algebra vector-spaces linear-transformations
linear-algebra vector-spaces linear-transformations
edited Jan 29 at 7:17
Chinnapparaj R
5,8602928
asked Jan 29 at 7:13
AbhayAbhay
3789
edited Jan 29 at 7:17
Chinnapparaj R
5,8602928
asked Jan 29 at 7:13
AbhayAbhay
3789
edited Jan 29 at 7:17
Chinnapparaj R
5,8602928
edited Jan 29 at 7:17
Chinnapparaj R
5,8602928
edited Jan 29 at 7:17
Chinnapparaj R
5,8602928
5,8602928
asked Jan 29 at 7:13
AbhayAbhay
3789
asked Jan 29 at 7:13
AbhayAbhay
3789
asked Jan 29 at 7:13
AbhayAbhay
3789
3789
$begingroup$
See proposition $5.21$ on page no:$101$ in this book!
$endgroup$
– Chinnapparaj R
Jan 29 at 7:36
add a comment |
$begingroup$
See proposition $5.21$ on page no:$101$ in this book!
$endgroup$
– Chinnapparaj R
Jan 29 at 7:36
$begingroup$
See proposition $5.21$ on page no:$101$ in this book!
$endgroup$
– Chinnapparaj R
Jan 29 at 7:36
$begingroup$
See proposition $5.21$ on page no:$101$ in this book!
$endgroup$
– Chinnapparaj R
Jan 29 at 7:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $T$ is similar to a diagonal matrix, there exists a basis $mathfrak{B}={v_j}_{jle dim V}$ of $V$ such that for each $j$, $v_j in W_k =ker (T-c_kI)$ for some $k$. Since $mathfrak{B} subset bigoplus_k W_k$, it follows
$$
V =text{span} mathfrak{B} =bigoplus_k W_k
$$ and $dim V =dim bigoplus_k W_k=sum_k dim W_k.$
answered Jan 29 at 7:35
SongSong
18.5k21651
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Since $T$ is similar to a diagonal matrix, there exists a basis $mathfrak{B}={v_j}_{jle dim V}$ of $V$ such that for each $j$, $v_j in W_k =ker (T-c_kI)$ for some $k$. Since $mathfrak{B} subset bigoplus_k W_k$, it follows
$$
V =text{span} mathfrak{B} =bigoplus_k W_k
$$ and $dim V =dim bigoplus_k W_k=sum_k dim W_k.$
answered Jan 29 at 7:35
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
Since $T$ is similar to a diagonal matrix, there exists a basis $mathfrak{B}={v_j}_{jle dim V}$ of $V$ such that for each $j$, $v_j in W_k =ker (T-c_kI)$ for some $k$. Since $mathfrak{B} subset bigoplus_k W_k$, it follows
$$
V =text{span} mathfrak{B} =bigoplus_k W_k
$$ and $dim V =dim bigoplus_k W_k=sum_k dim W_k.$
answered Jan 29 at 7:35
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
Since $T$ is similar to a diagonal matrix, there exists a basis $mathfrak{B}={v_j}_{jle dim V}$ of $V$ such that for each $j$, $v_j in W_k =ker (T-c_kI)$ for some $k$. Since $mathfrak{B} subset bigoplus_k W_k$, it follows
$$
V =text{span} mathfrak{B} =bigoplus_k W_k
$$ and $dim V =dim bigoplus_k W_k=sum_k dim W_k.$
answered Jan 29 at 7:35
SongSong
18.5k21651
$endgroup$
Since $T$ is similar to a diagonal matrix, there exists a basis $mathfrak{B}={v_j}_{jle dim V}$ of $V$ such that for each $j$, $v_j in W_k =ker (T-c_kI)$ for some $k$. Since $mathfrak{B} subset bigoplus_k W_k$, it follows
$$
V =text{span} mathfrak{B} =bigoplus_k W_k
$$ and $dim V =dim bigoplus_k W_k=sum_k dim W_k.$
answered Jan 29 at 7:35
SongSong
18.5k21651
answered Jan 29 at 7:35
SongSong
18.5k21651
answered Jan 29 at 7:35
SongSong
18.5k21651
answered Jan 29 at 7:35
SongSong
18.5k21651
18.5k21651
add a comment |
add a comment |
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$begingroup$
See proposition $5.21$ on page no:$101$ in this book!
$endgroup$
– Chinnapparaj R
Jan 29 at 7:36